In a senior secondary school, 80 students play hockey or football. The number that play football is 5 more than twice the number that play hockey. If the 15 students play both games and every student in the school playsat least one game, find:

Question

The number of students that play football;
The number of students that play football but not hockey;
The number of students that play hockey but not footbal.

Answer 1

number who play hockey ---- x
number who play football -- 2x+5

x + 2x+5 = 80
x = 25

so 25 play hockey, 55 play football, and 15 play both

now we can use a Venn diagram, filling in as follows:
15 in the intersection
55-15 or 40 in the football only part of the circle
25-15 or 10 in the hockey only part of the circle

All your questions can now be easily answered.

Answer 2

solve it for me

Answer 3

pls i really need the working and answer,can u help me with it

Answer 4

I don't get what reiny is trying to say

Answer 5

In a senior secondary school, 80 students play hockey or football. The number that play football is 5 more than twice the number that play hockey. If the 15 students play both games and every student in the school playsat least one game, find:
The number of students that play football;
The number of students that play football but not hockey;
The number of students that play hockey but not football. Full working sir

Answer 6

U={80}
5 more than twice of the number of hockey players plays football.
Let the hockey players(H) be X
Football players(F) be 2X+5
To find X
2X+5=FnH'+15. Equation1 X=HnF'+15. Equation2
80=(2X+5-15)+15+(X-15)
X=30

From here substitute X into equation1 and 2 to get F only nd H only

And remember that F will become 2(30)+5=65

Answer 7

In the department of mathematics, unilag, 80 students play hockey or football. The number that play football is 5 more than twice the number that play hockey. If 15 students play both games and every student in school plates at least one game. Find the number of students that play football but not hockey

Answer 8

H=x
F=5+2x
Therefore, x+5+2x+15=80
3x+20=80
3x=60
X=20

Proof: x+5+2x+15
20+5+40+15=80

Answer 9

I got lost on Alexander explanation starting from how to get x

Answer 10

this maths is confusing especially on how to get the number of those who played football when hockey is 5 and hockey only is 20. i.e H=5+2x+15+x=80. 5+15+2x+x=80 20+3x=80 3x=80-20 3x=60 hockey only=20 hockey=20-15 =5

Answer 11

D response I'm getin isn't helpin .d ansa from alexandra nd anonymous ren't same.wich should I chose x=20 or x=30

Answer 12

Equation on number of students that play football only

Answer 13

To get x = 30 just add the number of two games and the no that play both and equate it to no of the student, and no of the student is 80
X-15+15+(2x+5)-15=80
X+2x+5-15=80
3x-10=80
3x=80+10
3x=90
X=30

To find no. (ii) and (iii)
ii. n(F)only= total no of F-- none
And total no of F is 2x+5
=2(30)+5
=65
Therefore, n(F)only=65-15 =50

iii. n(H)only = total no of H -- none
And total no of H is x which is 30
Therefore, n(H)only=30-15
=15

Answer 14

Total number of student =80
N(F n H)=15
N(F)=5+2x
N(H)=x
(I). The number that play football
To get that , we need to sum up all the n( F) and n(H) to 80
I.e 5+2x-15 +15+X-15=80
3x-10=80
3x=90
X=30
Therefore; n(F)=5+2x
=5+60=65.
(ii). The number that play football but not hockey
N(F n H')
To get that we need to subtract the intercept from the to ta number that play football .and the intercept is 15
Therefore; 65-15=50
(ii). The number that play hockey but not football
N(F' n H)
We already know that x =30 and the intercept is 15
Therefore ; 30-15=15.

Answer 15

The number of those that played football but not hokey

Answer 16

This was so helpful

Answer 17

Thein a senior secondary school,80 students play hockey or football. The number that play football is 5 more than twice the number that play hockey. If 15 students play both games and every student in the school plays at least one game with the aid of using venn diagram :the number of students that play football picture of the Venn diagram for solving

Answer 18

The Venn diagram of the above question